Problem
ALG-B1-M10-P017 Three neighboring values
#17
★★★★☆ Level 4 of 5
Find all integer triples \((x,y,z)\) such that \(x^2+y^2+z^2=xy+yz+zx+3\).
Multiply by \(2\) and use the sum of squared differences.
We get \((x-y)^2+(y-z)^2+(z-x)^2=6\). The integer squares must be \(1,1,4\). Therefore the differences are \(1,1,-2\), up to signs and order. Hence the triple consists of three consecutive integers. All solutions are permutations of \((t-1,t,t+1)\), \(t\in\mathbb Z\).
A good mixed problem on symmetry and integer squares.